How do you solve #ln x - ln(x-1) = 2#?

1 Answer
Apr 4, 2018

Answer:

#x=e^2/(e^2-1)#

Explanation:

First note that

#lnx-ln(x-1)=ln(x/(x-1))#

So for your problem we can write

#ln(x/(x-1))=2#.

Now take the exponential of both sides

#x/(x-1)=e^2#

(Another way to think of this is if #lna=b#, then #a=e^b#.)

#x=xe^2-e^2#

#x=e^2/(e^2-1)=1/(1-1/e^2)#